Question 986773
<pre>
{{{ln(y)=5*ln(2x)}}}

Move the 5 coefficient of the logarithm to an exponent of what
the logarithm is taken of:

{{{ln(y)=ln(2x)^5}}}

As long as y and x are both positive, then we can
simply equate what the logs are taken of:

{{{y=(2x)^5}}} or

{{{y=32x^5}}}

Edwin</pre>